Chapter 39, Problem 74AP

(a)

To determine

The equation for $u$ from the given equation.

Answer to Problem 74AP

The equation for $u$ from the given equation is $c\sqrt{\frac{{\left(\frac{K}{m{c}^2}\right)}^2+2\left(\frac{K}{m{c}^2}\right)}{\left({\left(\frac{K}{m{c}^2}\right)}^2+2\left(\frac{K}{m{c}^2}\right)+1\right)}}$.

Explanation of Solution

Given info: The given equation is $K=\left(\frac{1}{\sqrt{1-u^2/c^2}}-1\right)m{c}^2$.

The equation for the kinetic energy is given as,

$K=\left(\frac{1}{\sqrt{1-u^2/c^2}}-1\right)m{c}^2$

Here,

$k$ is the kinetic energy.

$m$ is the mass.

$u$ is the speed.

$c$ is the speed of light.

Rearrange the above equation for $u$.

$\frac{K}{m{c}^2}=\left(\frac{1}{\sqrt{1-u^2/c^2}}-1\right)$

Let us assume $H=\frac{K}{m{c}^2}$.

$\begin{array}{c}H=\left(\frac{1}{\sqrt{1-u^2/c^2}}-1\right)\\ \frac{1}{\sqrt{1-u^2/c^2}}=1+H\\ 1-u^2/c^2={\left(\frac{1}{1+H}\right)}^2\\ u^2=c^2\left(1-{\left(\frac{1}{1+H}\right)}^2\right)\end{array}$

Further solve the equation.

$\begin{array}{c}{u}^2=c^2\left(1-{\left(\frac{1}{1+H}\right)}^2\right)\\ u^2=c^2\left(\frac{H^2+2H+1-1}{\left(H^2+2H+1\right)}\right)\\ u=c\sqrt{\frac{H^2+2H}{\left(H^2+2H+1\right)}}\end{array}$ (1)

Replace $H$ by $\frac{K}{m{c}^2}$ in above equation.

$u=c\sqrt{\frac{{\left(\frac{K}{m{c}^2}\right)}^2+2\left(\frac{K}{m{c}^2}\right)}{\left({\left(\frac{K}{m{c}^2}\right)}^2+2\left(\frac{K}{m{c}^2}\right)+1\right)}}$ (2)

Conclusion:

Therefore, the equation for $u$ from the given equation is $c\sqrt{\frac{{\left(\frac{K}{m{c}^2}\right)}^2+2\left(\frac{K}{m{c}^2}\right)}{\left({\left(\frac{K}{m{c}^2}\right)}^2+2\left(\frac{K}{m{c}^2}\right)+1\right)}}$.

(b)

To determine

The minimum possible value of speed and corresponding kinetic energy.

Answer to Problem 74AP

The minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

Explanation of Solution

Given info: The given equation is $K=\left(\frac{1}{\sqrt{1-u^2/c^2}}-1\right)m{c}^2$.

From equation (2), the expression for the speed is given as,

$u=c\sqrt{\frac{{\left(\frac{K}{m{c}^2}\right)}^2+2\left(\frac{K}{m{c}^2}\right)}{\left({\left(\frac{K}{m{c}^2}\right)}^2+2\left(\frac{K}{m{c}^2}\right)+1\right)}}$

From the above expression all the term is positive as well as the expression contains only positive sign so the minimum possible value that the speed can have according to the above expression is zero.

At zero speed the corresponding value of kinetic energy is also zero.

Conclusion:

Therefore, the minimum possible value of speed can be zero and corresponding kinetic energy will also be zero.

(c)

To determine

The maximum possible value of speed and corresponding kinetic energy.

Answer to Problem 74AP

The maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

Explanation of Solution

Given info: The given equation is $K=\left(\frac{1}{\sqrt{1-u^2/c^2}}-1\right)m{c}^2$.

From equation (2), the expression for the speed is given as,

$u=c\sqrt{\frac{{\left(\frac{K}{m{c}^2}\right)}^2+2\left(\frac{K}{m{c}^2}\right)}{\left({\left(\frac{K}{m{c}^2}\right)}^2+2\left(\frac{K}{m{c}^2}\right)+1\right)}}$

The maximum value of speed is equal to the speed of light according to relativistic concept if the speed becomes more than the speed of light then its energy become unstable that would not exist practically.

At this speed of light, the kinetic energy increases without any limit.

Conclusion:

Therefore, the maximum possible value of speed can be speed of light and corresponding kinetic energy will increases without any limit.

(d)

To determine

The equation for the acceleration of the particle as a function of kinetic energy and power input.

Answer to Problem 74AP

The equation for the acceleration of the particle as a function of kinetic energy and power input is $\frac{P}{mc{\left(\frac{K}{m{c}^2}\right)}^{1/2}{\left(\frac{K}{m{c}^2}+2\right)}^{1/2}{\left(\frac{K}{m{c}^2}+1\right)}^2}$.

Explanation of Solution

Given info: The given equation is $K=\left(\frac{1}{\sqrt{1-u^2/c^2}}-1\right)m{c}^2$.

From equation (1), the expression for the speed is given as,

$u=c\sqrt{\frac{{\left(H\right)}^2+2\left(H\right)}{\left({\left(H\right)}^2+2\left(H\right)+1\right)}}$

Write the expression for the acceleration of a particle.

$a=\frac{d^{2}u}{d{t}^2}$

Substitute $c\sqrt{\frac{{\left(H\right)}^2+2\left(H\right)}{\left({\left(H\right)}^2+2\left(H\right)+1\right)}}$ for $u$ to find $a$.

$\begin{array}{c}a=\frac{d^2\left(c\sqrt{\frac{{\left(H\right)}^2+2\left(H\right)}{\left({\left(H\right)}^2+2\left(H\right)+1\right)}}\right)}{d{t}^2}\\ =c{\left(\frac{{\left(H\right)}^2+2\left(H\right)}{\left({\left(H\right)}^2+2\left(H\right)+1\right)}\right)}^{-1/2}\left(\frac{H+1}{{\left(H+1\right)}^4}\right)\frac{d\left(H\right)}{dt}\\ =\frac{c}{H^{1/2}{\left(H+2\right)}^{1/2}{\left(H+1\right)}^2}\frac{d\left(H\right)}{dt}\end{array}$

Replace $H$ by $\frac{K}{m{c}^2}$ and $P$ by $\frac{dK}{dt}$.

$\begin{array}{c}a=\frac{c}{H^{1/2}{\left(H+2\right)}^{1/2}{\left(H+1\right)}^2}\frac{d\left(\frac{K}{m{c}^2}\right)}{dt}\\ =\frac{1}{mc{H}^{1/2}{\left(H+2\right)}^{1/2}{\left(H+1\right)}^2}\frac{dK}{dt}\\ =\frac{P}{mc{H}^{1/2}{\left(H+2\right)}^{1/2}{\left(H+1\right)}^2}\end{array}$ (3)

Substitute $\frac{K}{m{c}^2}$ for $H$ in above equation.

$a=\frac{P}{mc{\left(\frac{K}{m{c}^2}\right)}^{1/2}{\left(\frac{K}{m{c}^2}+2\right)}^{1/2}{\left(\frac{K}{m{c}^2}+1\right)}^2}$ (4)

Conclusion:

Therefore, the equation for the acceleration of the particle as a function of kinetic energy and power input is $\frac{P}{mc{\left(\frac{K}{m{c}^2}\right)}^{1/2}{\left(\frac{K}{m{c}^2}+2\right)}^{1/2}{\left(\frac{K}{m{c}^2}+1\right)}^2}$.

(e)

To determine

The limiting form of the expression in part (d) at low energy and compare with the non-relativistic expression.

Answer to Problem 74AP

The limiting form of the expression of acceleration at low energy is $\frac{P}{{\left(2mK\right)}^{1/2}}$ and it is same as that of expression of acceleration of non-relativistic case.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/{\left(2mK\right)}^{1/2}$.

From equation (4), the expression for the acceleration is given as,

$a=\frac{P}{mc{\left(\frac{K}{m{c}^2}\right)}^{1/2}{\left(\frac{K}{m{c}^2}+2\right)}^{1/2}{\left(\frac{K}{m{c}^2}+1\right)}^2}$

At low energy the value of $\frac{K}{m{c}^2}$ is very small that $\frac{K}{m{c}^2}<<<1$. So equation becomes,

$\begin{array}{c}a=\frac{P}{mc{\left(\frac{K}{m{c}^2}\right)}^{1/2}{\left(2\right)}^{1/2}{\left(1\right)}^2}\\ =\frac{P}{{\left(2mK\right)}^{1/2}}\end{array}$

Thus, the limiting form of the expression of acceleration at low energy is $\frac{P}{{\left(2mK\right)}^{1/2}}$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at low energy is $\frac{P}{{\left(2mK\right)}^{1/2}}$ and it is same as that of expression of acceleration of non-relativistic case.

(f)

To determine

The limiting form of the expression in part (d) at high energy and compare with the non-relativistic expression.

Answer to Problem 74AP

The limiting form of the expression of acceleration at high energy is $\frac{P{m}^2{c}^5}{{\left(K\right)}^3}$.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/{\left(2mK\right)}^{1/2}$.

From equation (4), the expression for the acceleration is given as,

$a=\frac{P}{mc{\left(\frac{K}{m{c}^2}\right)}^{1/2}{\left(\frac{K}{m{c}^2}+2\right)}^{1/2}{\left(\frac{K}{m{c}^2}+1\right)}^2}$

At high energy the value of $\frac{K}{m{c}^2}$ is very small that $\frac{K}{m{c}^2}>>>1$. So equation becomes,

$\begin{array}{c}a=\frac{P}{mc{\left(\frac{K}{m{c}^2}\right)}^{1/2}{\left(\frac{K}{m{c}^2}\right)}^{1/2}{\left(\frac{K}{m{c}^2}\right)}^2}\\ =\frac{P}{mc{\left(\frac{K}{m{c}^2}\right)}^3}\\ =\frac{P{m}^2{c}^5}{{\left(K\right)}^3}\end{array}$

Thus, the limiting form of the expression of acceleration at low energy is $\frac{P}{{\left(2mK\right)}^{1/2}}$ and it is same as that of expression of acceleration of non-relativistic case.

Conclusion:

Therefore, the limiting form of the expression of acceleration at high energy is $\frac{P{m}^2{c}^5}{{\left(K\right)}^3}$.

(g)

To determine

The reason that answer to part (f) help account for the answer to part (c) at constant input power.

Answer to Problem 74AP

The acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.

Explanation of Solution

Given info: The non-relativistic expression for the acceleration is $a=P/{\left(2mK\right)}^{1/2}$.

From the answer of part (f) the expression for the acceleration is,

$a=\frac{P{m}^2{c}^5}{{\left(K\right)}^3}$

Here,

$p$ is the power.

$m$ is the mass of a particle.

$K$ is the kinetic energy of the particle.

In part (c), the speed at high energy approaches to the speed of light. But from the acceleration equation if the energy is imparted to the particle at constant input power the acceleration is steeply decreases because the acceleration is inversely proportional to the cube root of the kinetic energy. So at high energy acceleration is very less and the velocity of the particle approaches to a constant value as indicate in part (c).

Conclusion:

Therefore, the acceleration of the particle is very less at high energy that gives the velocity of the particle a constant value.